Predict voltage from unequal ion concentrations

The same metal can have different electrode conditions

Place a pure copper electrode in each of two copper(II) solutions. Their standard reduction potentials are identical, so their difference E°cell is zero. That statement compares reference states. It does not say that the two actual electrodes have the same potential when their solution compositions differ.

Start with left exponent −3 and right exponent −1. These mean 0.001 mol/L on the left and 0.1 mol/L on the right. Each slider step changes its concentration by a factor of ten. An exponent of zero means 1 mol/L, not zero concentration. Zero concentration is excluded: the logarithm used below would not be defined there.

The model uses an ideal solution approximation and a fixed temperature of 298.15 K, or 25 °C. It estimates reversible cell potential, not current through a particular resistor. The ion connection is assumed present and ideal. A real measurement can include additional effects from nonideal activities, the connection between solutions, temperature differences and electrode condition.

Fix the direction before building the quotient

Keep the calculation convention fixed: left oxidation and right reduction. The half-reactions for that written direction are:

• Left: Cu(s, left) → Cu²⁺(left) + 2e−.
• Right: Cu²⁺(right) + 2e− → Cu(s, right).

Copper metal disappears on the left and is deposited on the right for a favorable forward reaction. The net solution change produces left-side copper ions while consuming right-side copper ions. Compartment labels matter: chemically identical ions in separate conditions cannot simply be canceled out of the problem.

For this written reaction, the dimensionless reaction quotient is Q = a_left / a_right. The symbol a means activity, a measure of effective concentration relative to a reference. In this model a is approximated by c divided by 1 mol/L, where c is concentration. Pure solid copper has activity one and contributes no changing factor. The quotient is therefore the left concentration divided by the right concentration when both use the same units.

Here Q is not the charge in coulombs used in the preceding lesson. The letter is reused in chemistry; its definition and units determine its meaning. This Q is dimensionless and describes composition, not transferred electrons.

Use the logarithm as a count of factors of ten

At the initial settings, Q = 0.001 / 0.1 = 0.01. Since 0.01 = 10⁻², log₁₀ Q = −2. You can obtain that value directly by subtracting the right exponent from the left: −3 − (−1) = −2.

The Nernst relation is E = E° − (RT/nF) ln Q. E is the potential for the written direction, E° its standard potential, R the gas constant, T absolute temperature in kelvin, n electron stoichiometry and F charge per mole of electrons. ln denotes the natural logarithm. Here E° = 0 V and n = 2. We use rounded constants R = 8.314 J/(mol K) and F = 96485 C/mol electrons.

Because ln Q = ln(10) × log₁₀ Q, the same calculation becomes E ≈ −0.029578 × log₁₀ Q volts at this temperature. For Q = 0.01, E is about +0.059156 V. The widget retains more arithmetic precision internally than this worked example; its displayed digits do not imply measurement precision.

The OpenStax treatment of nonstandard potentials and concentration cells supplies the underlying Nernst relation. The settings, fixed scales and paired comparisons in this investigation are constructed teaching cases.

Predict the direction and then reverse it

Positive right-minus-left potential means the chosen left-oxidation/right-reduction direction is favorable. On closing a suitable external circuit, the left electrode supplies electrons and the right receives them. Oxidation increases the dilute left solution’s copper-ion amount; reduction decreases the more concentrated right solution’s amount. This reduces the concentration difference if compartment volumes stay fixed and other transfers are neglected.

Now swap the exponents to left −1 and right −3. Q becomes 100, log₁₀ Q becomes +2, and E becomes approximately −0.059156 V under the same fixed voltage convention. The favorable reaction reverses: right copper oxidizes and left copper ions reduce. The widget changes the anode/cathode labels to describe that favorable direction, while its calculation table keeps the original written convention. A negative displayed potential is not a negative electron count.

Compare the two captures. Their potentials have opposite signs but equal magnitudes. Explain the change from composition, not by renaming the electrodes until the sign disappears.

Hold the ratio constant in a second comparison

Compare exponent pairs (−4, −2) and (−3, −1). Both have Q = 0.01 and therefore the same ideal potential. Their individual concentrations differ by a factor of ten. The ratio determines this model’s potential; a larger concentration difference in mol/L does not by itself establish a larger voltage.

That does not establish equal available charge, equal resistance or equal power. Those quantities require inventories, volumes, transport or time information that the model does not provide. At high concentrations the ideal activity approximation may also fail. The controls deliberately isolate one thermodynamic relation.

Recognize the equal-activity condition

Set both exponents to −2. Q = 1, ln Q = 0 and E = 0. The nonstandard reaction free energy ΔrG = −2FE is also zero for this reaction at that composition. Equal exponents −4/−4 or 0/0 give the same no-net-driving-force result. Equilibrium here requires equal activities, not necessarily that each solution is in its standard state.

This does not imply that all ions vanish or that microscopic electrode exchange stops. It is a statement about net thermodynamic drive. The controls do not evolve a conserved initial inventory toward equilibrium: changing a slider replaces a composition snapshot. To calculate a final concentration after discharge, you would also need volumes, initial amounts and appropriate material balances. Do not read a sequence of slider choices as measured time data.

Transfer: choose left concentration 0.0001 mol/L and right 0.1 mol/L. Calculate Q, log₁₀ Q, the potential sign and approximate magnitude. Then explain what changes if both concentrations are multiplied by ten while remaining within the model range.

Check the concentration-cell transfer

Q = 0.001 and log₁₀ Q = −3, so E ≈ +0.088734 V. Left oxidation and right reduction are favorable. Multiplying both concentrations by ten leaves Q unchanged, so the ideal potential is unchanged. It changes the amounts present for fixed volumes and may change real solution behavior, which this calculation does not predict.